When is a knowledge base consistent?
I will first recapitulate the key concepts which you need to know in order to understand the answer to your question (which will be very simple, because I will just try to clarify what is given as a &quot;definition&quot;).
In logic, a formula is e.g. $f$, $\lnot f$, $f \land g$, where $f$ can be e.g. the proposition (or variable) &quot;today it will rain&quot;.  So, in a (propositional) formula, you have propositions, i.e. sentences like &quot;today it will rain&quot;, and logical connectives, i.e. symbols like $\land$ (i.e. logical AND), which logically connect these sentences. The propositions like &quot;today it will rain&quot; can often be denoted by a single (capital) letter like $P$. $f \land g$ is the combination of two formulae (where formulae is the plural of formula). So, for example, suppose that $f$ is composed of the propositions &quot;today it will rain&quot; (denoted by $P$) or &quot;my friend will visit me&quot; (denoted by $Q$) and $f$ is defined as &quot;I will play with my friend&quot; (denoted by $S$). Then the formula $f \land g = (P \lor Q) \land S$. In general, you can combine formulae in any logically appropriate way.
In this context, a model is an assignment to each variable in a formula. For example, suppose $f = P \lor Q$, then $w = \{ P=0, Q = 1\}$ is a model for $f$, that is, each variable (e.g. $P$) is assigned either &quot;true&quot; ($1$) or &quot;false&quot; ($0$) but not both. (Note that the word model may be used to refer to different concepts depending on the context; again, in this context, you can simply think of a model as an assignment of values to the variables in a formula.)
Suppose now we define $I(f, w)$ to be a function that receives the formula $f$ and the model $w$ as input, and $I$ returns either &quot;true&quot; ($1$) or &quot;false&quot; ($0$). In other words, $I$ is a function that automatically tells us if $f$ is evaluated to true or false given the assignment $w$.
You can now define $M(f)$ to be a set of assignments (or models) to the formula $f$ such that $f$ is true. So, $M$ is a set and not just an assignment (or model). This set can be empty, it can contain one assignment or it can contain any number of assignments: it depends on the formula $f$: in some cases, $M$ is empty and, in other cases, it may contain say $n$ valid assignments to $f$, where by &quot;valid&quot; I mean that these assignments make $f$ evaluate to &quot;true&quot;. For example, suppose we have formula $f = A \land \lnot A$. Then you can try to assign any value to $A$, but $f$ will never evaluate to true. In that case, $M(f)$ is an empty set, because there is no assignment to the variables (or propositions) of $f$ which make $f$ evaluate to true.
A knowledge base is a set of formulae $\text{KB} = \{ f_1, f_2, \dots, f_n \}$. So, for example, $f_2 = $ &quot;today it will rain&quot; and $f_3 = $ &quot;I will go to school AND I will have lunch&quot;.
We can now define $M(\text{KB})$ to be the set of assignments to the formulae in the knowledge base $\text{KB}$ such that all formulae are true. If you think of the formulae in $KB$ as &quot;facts&quot;, $M(\text{KB})$ is an assignment to these formulae in $KB$ such that these facts hold or are true.
In this context, we then say that a particular knowledge base (i.e., a set of formulae as defined above), denoted by $\text{KB}$, is consistent with formula $f$ if $M(\text{KB} \cup \{ f \})$ is a non-empty set, where $\cup$ means the union operation between sets: note that (as we defined it above) $\text{KB}$ is a set, and $\{ f \}$ means that we are making a set out of the formula $f$, so we are indeed performing an union operation on sets.
So, what does it mean for a knowledge base to be consistent? First of all, the consistency of a knowledge base $\text{KB}$ is defined with respect to another formula $f$. Recall that a knowledge base is a set of formulae, so we are defining the consistency of a set of formulae with respect to another formula.
When is then a knowledge base $\text{KB}$ consistent with a formula $f$? When $M(\text{KB} \cup \{ f \})$ is a non-empty set. Recall that $M$ is an assignment to the variables in its input such that its inputs evaluate to true. So, $\text{KB}$ is consistent with $f$ when there is a set of assignments of values to the formulae in $\text{KB}$ and an assignment of values to the variables in $f$ such that both $\text{KB}$ and $f$ are true. In other words, $\text{KB}$ is consistent with $f$ when both all formulae in $\text{KB}$ and $f$ can be true at the same time.